The angles in triangle ABC satisfy 6sin∠A=3√(3)sin∠B=2√(2)sin∠C.
If sin^2∠A=a/b, where a and b are coprime positive integers, what is the value of a+b?
To find the value of a+b, we need to determine the value of sin^2∠A first.
Given that the angles in triangle ABC satisfy 6sin∠A = 3√(3)sin∠B = 2√(2)sin∠C, we can simplify these ratios by dividing each ratio by their respective coefficients.
sin∠A = (3√(3))/6 = √(3)/2
sin∠B = (2√(2))/6 = √(2)/3
sin∠C = (2√(2))/6 = √(2)/3
Since sin^2∠A = (sin∠A)^2, we can square the value of sin∠A.
(sin∠A)^2 = (√(3)/2)^2 = (3/4)
Now, we know that sin^2∠A = 3/4. To express this as a fraction where a and b are coprime positive integers, we can write it as 3/4 = a/b.
Since a = 3 and b = 4 are coprime positive integers, the value of a+b is 3 + 4 = 7.
Therefore, the value of a+b is 7.